What is the solution for y^{''}-5y^'=x-2 ?

The solution for y^{''}-5y^'=x-2 is y=c_{1}e^{5x}+c_{2}-(x^2)/(10)+(9x)/(25)

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y^{''}-5y^'=x-2

y^{^{''}} - 5 y^{^{'}} = x - 2

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{2}}{10} + \frac{9 x}{25}

Solution steps

y^{''} (x) - 5 y^{'} (x) = x - 2

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{2}}{10} + \frac{9 x}{25}

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{2}}{10} + \frac{9 x}{25}

3 x^{2} cos (x) cot (x)

e^{x} sin (2 x)

\frac{d}{d x} (ln (\frac{x}{k}))

\int_{- a}^{a} \frac{1}{( s - x ) ^{2}} d x

\frac{d}{d x} (\frac{x ^{2} - 4}{x ^{2} - 2 x})

What is the solution for y^{''}-5y^'=x-2 ?
The solution for y^{''}-5y^'=x-2 is y=c_{1}e^{5x}+c_{2}-(x^2)/(10)+(9x)/(25)